How many liters of antifreeze ethylene glycol[CH 2 (OH)CH 2 (OH)] would you add

Question

How many liters of antifreeze ethylene glycol[CH2(OH)CH2(OH)] would you add to a carradiator containing 7.50 L of water if the coldest wintertemperature in your area is -19.5 degree C? calculate the boilingpoint of this water-ethylene glycol mixture. (The density ofethylene glycol is 1.11 g/mL) a) what is the volume of antifreeze (in L)? b) what is the boiling point of the solution ? How many liters of antifreeze ethylene glycol[CH2(OH)CH2(OH)] would you add to a carradiator containing 7.50 L of water if the coldest wintertemperature in your area is -19.5 degree C? calculate the boilingpoint of this water-ethylene glycol mixture. (The density ofethylene glycol is 1.11 g/mL) a) what is the volume of antifreeze (in L)? b) what is the boiling point of the solution ?

Explanation / Answer

a) This could be solved by considering the colligativeproperties .     Tf = Kf * m Where Tf is the depression in freezing point            Kf is the molal depression in freezing constant            m is the molality Tf   = Tpure - Tsolution       = 0 0C - ( -19.5 0 C)         =19.5 0 C    Kf for water   =1.86  0 C / m m   =  19.5 0 C /1.86  0 C / m           = 10.48 m since the density of water is 1g / mL , mass of solvent = 7.50 kg molality = No.of moles of solute / Mass of solvent   No.of moles of solute   = 10.48 m / 7.50 kg                                     =   1.39 moles Mass of etylene glycol    =   1.39moles * 62.068 g/mol
                                     =   86.2745 g Given that the density of ethylene glycol  = 1.11 g/ mL Volume of ethylene glycol = 86.2745 g /1.11 g / mL                                             = 0.077 L b )  Tb = Kb * m      Where Tb is the elevation in boilingpoint            Kb is the molal elevation in boiling constant ( 0.510 C / m for water )            m is the molality    Tb = 0.51 0 C / m*  10.48 m               = 5.3448  0 C                                                       Where Tb is the elevation in boilingpoint            Kb is the molal elevation in boiling constant ( 0.510 C / m for water )            m is the molality    Tb = 0.51 0 C / m*  10.48 m               = 5.3448  0 C                                                       But Tb = Tsoln -T solvent 5.3448 0C =  T solution-1000C     T solution  = 1000C +5.3448 0C                      =105.3448  0C          

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